Solution
= Solution
The <Maclaurin series> $\sin x=x-x^3/6+O(x^5)$ cancels the numerator's linear term, leaving
$$
\frac{\sin x-x}{x^3}=-\frac16+O(x^2).
$$
Thus $\boxed{\lim_{x\to0}(\sin x-x)/x^3=-1/6}$. The same expansion applies for positive and negative $x$, so this is a two-sided <limit>.